Death Star Energy Calculator

Select a target, set beam efficiency and pulse duration, then fire. The physics is real - the application is not.

Worked example — Death Star Energy Calculator

Required energy	6.40 × 10³² J
Required power	1.07 × 10³² W
Sun-output equivalent	19.4 d
Gravitational binding energy	(3/5) × G × M² / R

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Lesson

The theory — Death Star Energy Calculator

Blowing a planet apart means overcoming its own gravity, and the size of that job has a name: the gravitational binding energy. It is the energy you would have to supply to disassemble the body and carry every piece infinitely far away — for a uniform sphere, U = 3GM² / (5R).

What each symbol means

M: the planet’s mass. The default 5.972 × 10²⁴ kg is Earth’s, and it enters squared — every piece attracts every other, so the bill grows faster than the mass.
R: the radius, 6.371 × 10⁶ m. It divides, so a denser planet of the same mass is harder to break.
efficiency: the share of supplied energy that does useful work, 35% here. The requirement is the binding energy divided by this.
pulse: the pulse duration in seconds. Energy divided by time is power, which is where the absurd figure comes from.

Where the formula comes from

For a uniform sphere the binding energy is U = 3GM² / (5R). With Earth’s mass and radius that is about 2.24 × 10³² J.
Only 35% of what the weapon delivers does the work, so the energy it must supply is 2.24 × 10³² / 0.35 ≈ 6.40 × 10³² J — the figure reported above.
Divide by the pulse length to get power: 6.40 × 10³² / 6 ≈ 1.07 × 10³² W. For scale the page converts that to Sun-output: the Sun radiates 3.828 × 10²⁶ W, so this is 19.4 days of the entire Sun, released in six seconds.

How to read what you see

Three figures — required energy, required power, and the same energy expressed as days of total solar output. The third exists because the first two are past the point where scientific notation means anything to a human; 19.4 days of the Sun is a number you can actually feel.

Assumes: A uniform sphere, which Earth is not — it has a dense iron core, so its true binding energy is somewhat higher than the uniform formula gives. It also assumes the only job is undoing gravity, ignoring the energy needed to vaporise rock and the fact that most of it would radiate away rather than pushing anything apart.
Breaks when: The result is a lower bound, and it is already impossible. Nothing here explains where such a weapon stores 19 days of the Sun, or how it survives holding it — a beam carrying 10³² W would destroy its own emitter long before its target. The calculation is honest about the physics and silent about the engineering, which is the usual pattern when a film’s numbers are taken seriously: the energy is not the problem, the containment is.

Doubling the mass quadruples the bill 🖖

Why M² (not M) is the one that matters: gravitational binding energy comes from every particle in the sphere pulling on every other particle, so the interaction count itself scales with mass squared — double the mass and you roughly quadruple the energy needed to blow it apart, all else equal. R shows up in the denominator because spreading the same mass over a bigger sphere weakens its self-gravity; the 3/5 factor is just the geometry of integrating that pairwise pull over a uniform sphere instead of a point mass. Real numbers make the fiction obvious: Earth's own binding energy is about 2×10³² J — roughly 400 billion times humanity's total annual energy use — which is why a 'planet-destroying superweapon' stays squarely in science fiction no matter how efficient its beam is.

It's really an energy ladder 🖖

The tool's real job is to place a science-fiction feat onto a ladder of real energies — from the largest nuclear device ever tested (Tsar Bomba), up through all of humanity's yearly energy use, to the raw output of a star. The eye-opener: our Sun radiates enough energy to gravitationally unbind an Earth-sized planet roughly every week. So the quantity isn't cosmically rare — an ordinary star pours it out continuously. What's pure fiction is funnelling it into a single beam in seconds.

The Sun once 'ran' on this formula 🖖

The very quantity you're computing, (3/5)GM²/R, was once humanity's best guess for what powers the Sun. In the 1800s Kelvin and Helmholtz proposed it shines by slowly contracting, converting its own gravitational binding energy (~7×10⁴¹ J) into light. But dividing that by the Sun's luminosity gives only a few tens of millions of years of sunlight — far too short for the geological and fossil record. The paradox stood until nuclear fusion was discovered.

Problem solved in full

Energy required to disperse Earth so no piece falls back 5 steps
How much energy does it actually take to blow up a planet? Not to crack it — to disperse it, so no piece falls back. Work it out for Earth, then see what kind of power source that implies.
1. The target is the gravitational binding energy: the work needed to carry every shell of the planet out to infinity against the gravity of everything already inside it. Integrating shell by shell for a uniform sphere gives the 3/5, and real planets are denser at the centre, so this is an underestimate.
  
  $U = \int_{0}^{R} \frac{G m ( r )}{r} d m = \frac{3}{5} \frac{G M ^{2}}{R} (uniform density)$
2. Putting Earth's mass and radius in gives 2.24 × 10³² joules. Nothing about that number is exotic — it follows from G, a mass and a radius.
  
  $U = \frac{3}{5} \cdot \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.972 \times 1 0 ^{24} ) ^{2}}{6.371 \times 1 0 ^{6}} = 2.24 \times 1 0^{32} J$
3. No weapon converts stored energy to output perfectly. At 35% efficiency the machine must supply 6.40 × 10³².
  
  $η = 35% : E = \frac{2.24 \times 1 0 ^{32}}{0.35} = 6.40 \times 1 0^{32} J$
4. Delivered in a six-second shot, that is a power of 1.07 × 10³² watts.
  
  $Δ t = 6 s : P = \frac{6.40 \times 1 0 ^{32}}{6} = 1.07 \times 1 0^{32} W$
5. The Sun radiates 3.83 × 10²⁶ watts in every direction. So the beam runs at roughly 280 000 times the Sun's entire output, and the shot spends what the Sun emits in 19.4 days.
  
  $\frac{6.40 \times 1 0 ^{32}}{3.828 \times 1 0 ^{26}} = 1.67 \times 1 0^{6} s = 19.4 d of the Sun’s ENTIRE output$
Answer

The tool prints 6.40 × 10³² J, 1.07 × 10³² W and 19.4 days of total solar output. The physics is real even if the station is not, and the interesting part is which number is the problem. The energy is merely enormous; the power is the impossible bit, because it has to be delivered in seconds. Since binding energy goes as M²/R, a body twice Earth's mass and radius costs twice as much, not eight times — and Jupiter, at 318 masses but only 11 radii, costs about 9200 times as much. Switch the target and watch the exponent move.

$U = \frac{3}{5} \frac{G M ^{2}}{R} \propto \frac{M ^{2}}{R} : 19.4 days of total solar output$

References (1)

Earth's mass and radius and the Sun's luminosity, all three taken as IAU nominal values: A. Prša et al., "Nominal Values for Selected Solar and Planetary Quantities: IAU 2015 Resolution B3." The Astronomical Journal 152(2), 41, 2016.

By the 42Math team · Updated 2026-06-09

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Alderaan moment - Alderaan's real gravitational binding energy, delivered in 3 seconds at 35% efficiency, demands about 3.8×10³² watts — roughly a million times the Sun's entire power output, sustained for the whole pulse. That's why the 'planet-destroying superweapon' stays firmly in science fiction.
Warning shot - Even the Moon — small and low-density as planetary bodies go — needs about 2.5×10²⁹ joules to blow apart at 50% efficiency: roughly 1.2 trillion Tsar Bombas' worth of energy, the largest nuclear device ever detonated, delivered in one 10-second burst.
Slow charge - Stretch the firing time to two full minutes and Earth's gravitational binding energy still demands about 5.3×10³⁰ watts — nearly 14,000 times the Sun's entire power output. The total energy involved equals roughly 1.1 trillion years of humanity's current annual energy use.
Mars test - Mars is far easier to crack than Earth — less than a ninth the mass at a similar radius — but even at 40% efficiency it still takes about 2.0×10³⁰ watts, roughly 5,300 times the Sun's total power output, to do it in 6 seconds.